Question

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$. [CBSE 2015]

Answer 1

Let the equation of the plane passing through (3, 2, 0) be

$a\left(x-3\right)+b\left(y-2\right)+c\left(z-0\right)=0$ .....(1)

The line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ passes through the point (3, 6, 4) and its direction ratios are proportional to 1, 5, 4.

If plane (1) contains this line, then it must pass through (3, 6, 4) and must be parallel to the line.

$$\begin{gathered} \therefore a\left(3-3\right)+b\left(6-2\right)+c\left(4-0\right)=0 \\ \Rightarrow 4b+4c=0 \\ \Rightarrow b+c=0.....\left(2\right) \end{gathered}$$

Also,

$$\begin{gathered} 1\times a+5\times b+4\times c=0 \\ \Rightarrow a+5b+4c=0.....\left(3\right) \end{gathered}$$

Solving (2) and (3), we get

$$\begin{gathered} \frac{a}{4-5}=\frac{b}{1-0}=\frac{c}{0-1} \\ \Rightarrow \frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}=\lambda \left(\mathrm{Say}\right) \\ \Rightarrow a=-\lambda ,b=\lambda ,c=-\lambda \end{gathered}$$

Putting these values of a, b, c in (1), we get

$$\begin{gathered} -\lambda \left(x-3\right)+\lambda \left(y-2\right)-\lambda \left(z-0\right)=0 \\ \Rightarrow -x+3+y-2-z=0 \\ \Rightarrow -x+y-z+1=0 \\ \Rightarrow x-y+z-1=0 \end{gathered}$$

Thus, the equation of the required plane is x − y + z − 1 = 0.